Write out the form of the partial fraction decomposition of $\frac { x ^ { 2 } + 5 } { x ^ { 3 } - 2 x ^ { 2 } + 4 x - 8 }$
A) $\frac { A } { x - 2 } + \frac { B x } { ( x - 2 ) ^ { 2 } } + \frac { C } { x + 2 }$
B) $\frac { A x + B } { x - 2 } + \frac { C } { x ^ { 2 } + 4 }$
C) $\frac { A } { x - 2 } + \frac { B x } { ( x - 2 ) ^ { 2 } }$
D) $\frac { A } { x - 2 } + \frac { B x + C } { x ^ { 2 } + 4 }$
E)none

Question

Quizplus · Accepted Answer

The degree of the numerator is less than the degree of the denominator. Therefore, we can use partial fraction decomposition. The denominator can be factored as $(x-2)(x^2+4)$, so the partial fraction decomposition has the form $$\frac{x^2+5}{x^3-2x^2+4x-8}=\frac{A}{x-2}+\frac{Bx+C}{x^2+4}.$$ Multiplying both sides by the denominator gives  $$x^2+5=A(x^2+4)+(Bx+C)(x-2).$$Setting $x=2$ in this equation gives $9A=9$, so $A=1$. Substituting $x=0$ gives $5=-2A-2C$, so $C=-\frac{7}{2}$. Finally, comparing coefficients of $x$ in $x^2+5=A(x^2+4)+(Bx+C)(x-2)$ gives $0=4A+(B-2C)$. Substituting $A=1$ and $C=-\frac{7}{2}$ gives $B=\frac{1}{2}$. Therefore, the partial fraction decomposition is  $$\frac{x^2+5}{x^3-2x^2+4x-8}=\boxed{\frac{1}{x-2}+\frac{\frac{1}{2}x-\frac{7}{2}}{x^2+4}}.$$

Write Out the Form of the Partial Fraction Decomposition Of \frac